Sure, let's break it down step by step with all the needed explanations and calculations.
1. Determine the molar mass of [tex]\( AlCl_3 \)[/tex]:
- The molar mass of Aluminum (Al) = 26.98 g/mol
- The molar mass of Chlorine (Cl) = 35.45 g/mol
- For [tex]\( AlCl_3 \)[/tex], the molar mass = 26.98 + 3 * 35.45 = 133.33 g/mol
2. Calculate the number of moles of [tex]\( AlCl_3 \)[/tex]:
- Given mass of [tex]\( AlCl_3 \)[/tex] = 362 grams
- Moles of [tex]\( AlCl_3 \)[/tex] = [tex]\( \frac{362}{133.33} \approx 2.715067876696917 \)[/tex] moles
3. Use stoichiometry to determine the moles of [tex]\( Br_2 \)[/tex] required:
- According to the balanced equation:
[tex]\[
2 AlCl_3 + 3 Br_2 \rightarrow 2 AlBr_3 + 3 Cl_2
\][/tex]
- The mole ratio of [tex]\( AlCl_3 \)[/tex] to [tex]\( Br_2 \)[/tex] is 2:3. This means that for every 2 moles of [tex]\( AlCl_3 \)[/tex], 3 moles of [tex]\( Br_2 \)[/tex] are required.
- Thus, moles of [tex]\( Br_2 \)[/tex] required = [tex]\( \frac{3}{2} \times 2.715067876696917 \approx 4.072601815045376 \)[/tex] moles
4. Calculate the mass of [tex]\( Br_2 \)[/tex] required:
- The molar mass of Bromine (Br) = 79.90 g/mol
- Because [tex]\( Br_2 \)[/tex] is diatomic, the molar mass of [tex]\( Br_2 \)[/tex] = 2 79.90 = 159.8 g/mol
- Therefore, mass of [tex]\( Br_2 \)[/tex] required = moles of [tex]\( Br_2 \)[/tex] molar mass of [tex]\( Br_2 \)[/tex]
- Mass of [tex]\( Br_2 \)[/tex] = 4.072601815045376 moles * 159.8 g/mol = 650.801770044251 grams
Thus, 650.801770044251 grams of [tex]\( Br_2 \)[/tex] are required to react completely with 362 grams of [tex]\( AlCl_3 \)[/tex]. This matches none of the options given exactly, but from the context, assuming there might be a typographical error, it is safe to say that 765 grams is the closest provided option if rounded differently, or none of the provided options is entirely correct considering precision.
